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Viefleur [7K]
3 years ago
14

share 400 sweets between two boys such that one boy gets 60 more sweets than the other. what are the two shares?​

Mathematics
1 answer:
atroni [7]3 years ago
8 0

Answer:

170, 230

Step-by-step explanation:

Let one boy gets x sweets .

Other boy gets x + 60 sweets .

Total = 400

x + x + 60 = 400

2x = 340

x = 170

So the other gets = 170 + 60 = 230

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HACTEHA [7]

Answer:

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7 0
3 years ago
Which statement is true about the graph of two lines y=-6 and x=1/6
Gelneren [198K]
The graph of the line y = -6 is a horizontal line which passes through the y axis at -6. On the other hand, the graph of the equation, x = 1/6 is a vertical line that passes through the x-axis at 1/6. The intersection of these lines is at a point that is expressed as (1/6, -6). 
7 0
3 years ago
The lifespan (in days) of the common housefly is best modeled using a normal curve having mean 22 days and standard deviation 5.
Natasha_Volkova [10]

Answer:

Yes, it would be unusual.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If Z \leq -2 or Z \geq 2, the outcome X is considered unusual.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 22, \sigma = 5, n = 25, s = \frac{5}{\sqrt{25}} = 1

Would it be unusual for this sample mean to be less than 19 days?

We have to find Z when X = 19. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{19 - 22}{1}

Z = -3

Z = -3 \leq -2, so yes, the sample mean being less than 19 days would be considered an unusual outcome.

7 0
3 years ago
Four balls of wool will make 8 knitted caps. How many balls of wool will Malcolm need if he wants to make 6 caps
Maksim231197 [3]
4x2=8

3x2=6

so answer is 3
8 0
3 years ago
An office manager has received a report from a consultant that includes a section on equipment replacement. the report indicates
wolverine [178]

Answer:

a) 22.663%

b) 44%

c) 38.3%

Explanation:

An office manager has received a report from a consultant that includes a section on equipment replacement. The report indicates that scanners have a service life that is normally distributed with a mean of 41 months and a standard deviation of 4 months. On the basis of this information, determine the percentage of scanners that can be expected to fail in the following time periods:

We solve the above question using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean = 41 months

σ is the population standard deviation = 4 months

a. Before 38 months of service

Before in z score score means less than 38 months

Hence,

z = 38 - 41/4

z = -0.75

Probability value from Z-Table:

P(x<38) = 0.22663

Converting to percentage = 0.22663 × 100

= 22.663%

b. Between 40 and 45 months of service

For x = 40 months

z = 40 - 41/4

z = -0.2

Probaility value from Z-Table:

P(x =40) = 0.40129

For x = 45

z = 45 - 41/4

z = 1

Probability value from Z-Table:

P(x = 45) = 0.84134

Between 40 and 45 months of service

= 0.84134 - 0.40129

= 0.44005

Converting to Percentage

= 0.44005 × 100

= 44.005%

= 44%

c. Wihin ± 2 months of the mean life

+ 2 months = 41 months + 2 months

= 43 month

- 2 months = 41 months - 2 months

= 39 months

For x = 43

z = 43 - 41 /4

z = 0.5

P-value from Z-Table:

P(x = 43) = 0.69146

For x = 39

z = 39 - 41/4

z = -2/4

z = -0.5

Probability value from Z-Table:

P(x = 39) = 0.30854

Within ± 2 months of the mean life

= 0.69146 - 0.30854

= 0.38292

= 38.3%

Learn more about z-score:

brainly.com/question/17436641

#SPJ4

4 0
2 years ago
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